Question

If the circle $C_1:x^2+y^2=16$ intersects another circle $C_2$ of radius $5$ in such a manner that the common chord is of maximum length and has a slope equal to $\frac{3}{4}$, then the coordinates of the centre of $C_2$ are

• A. $(\pm \frac{9}{5},\pm \frac{12}{5})$
• B. $(\pm \frac{9}{5},\mp \frac{12}{5})$
• C. $(\pm \frac{12}{5},\pm \frac{9}{5})$
• D. $(\pm \frac{12}{5},\mp \frac{9}{5})$

Answer 1

Answer: (B)

$(\pm \frac{9}{5},\mp \frac{12}{5})$

$C_1:x^2+y^2=4^2$

$C_2:(x-h{)}^2+(y-k{)}^2=5^2$

Common chord is given by $C_2–C_1=0$

$⟹2xh+2yk+9=k^2+h^2$

Given slope $=\frac{3}{4}$

$\frac{-2h}{2k}=\frac{3}{4}⟹h=\frac{-3}{4}k$

The chord of maximum length is diameter.

$C_1(0,0)$ lies on common chord.

$⟹h^2+k^2=9$

$⟹\frac{9}{16}{k}^2+k^2=9⟹k=\pm \frac{12}{5}⟹h=\mp \frac{9}{5}$

$(h,k)=(\pm \frac{9}{5},\mp \frac{12}{5})$